# Eigenspinors of Sy (Griffiths 4.29)

1. Oct 31, 2011

### Rubiss

1. The problem statement, all variables and given/known data

Griffiths Quantum Mechanics, 2nd edition:
Calculate the eigenvalues and eigenspinors of Sy

2. Relevant equations

See attached PDF. Everything is in the PDF.

3. The attempt at a solution

See attached PDF. All of my steps are written out. I promise.

Parts 1, 2, and 3 are all in the attached PDF, I promise. I wrote out all of the steps in my work, so you can follow along. I am wondering where I made a math error in calculating the eigenspinors of Sy. I have some questions about b and c, and my work is shown for those parts as well. Hopefully someone can help.

I'm really sorry I couldn't write all of my math in this message board. I tried at first, but I couldn't get the matrices to work. So I wrote everything in my latex editor, and the result is the attached PDF.

Thanks.

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2. Nov 1, 2011

### vela

Staff Emeritus
Did you notice your equation (4) and your solution for $\alpha$ at the bottom of page 1 are the same?

Eigenvectors are unique only up to a multiplicative constant, so just arbitrarily set one of the variables to 1 and solve for the other and then normalize.

Your results are correct, but you have them swapped. If you multiply by $\pm i$, you'll get the answers you're looking for.

Your work for part (b) is fine.

Your work for part (c) isn't. You don't find probabilities by calculating the expectation value. You expand the state in terms of the eigenstates of the observable and then pick off the coefficients.

3. Nov 1, 2011

### Rubiss

Vela, thank you for responding. I greatly appreciate it. I didn't notice that equation (4) and my solution for $\alpha$ was the same. Thanks for pointing that out.

I not sure I understand what you are saying. Why can't I use my method of plugging alpha into my eigenspinor, and obtaining the correct result? I can use equation (4) to solve for $\beta$ and obtain $\beta=\pm i \alpha$. When I plug in these two values of $\beta$ into my eigenspinor, I obtain the correct result. Why should these way be any different?

I'm not sure I understand how my results are correct. Aren't the eigenspinors I obtained [equations (9) and (12)] fundamentally different from the correct eigenspinors [equations (13) and (14)]?

But is there any other way to solve this problem without using the identity operator?

Ok. From part b, I have expanded the state in terms of the eigenstates of the observable Sy. How do I expand the state in terms of the eigenstates of the observable Sy2?

4. Nov 1, 2011

### vela

Staff Emeritus
I was just pointing out that most of the work you did after getting equation (4) is unnecessary. Once you have (4), just set, say, $\alpha = 1$ and you'll get $\beta=\pm i$.
Take equation (9). Multiply it by -i. You get (14). Similarly, Multiply (12) by i, and you get (13).
If the state of the system is $| \psi \rangle$, the probability of finding the system in the state $| \phi \rangle$ is $|\langle \phi | \psi \rangle |^2$. This is true for any two states. When you use eigenstates for $|\phi\rangle$, you essentially get your results. Really, all you need are the coefficients in the expansion, so that's all you need to calculate.
What are the eigenstates of Sy2 and their associated eigenvalues?

5. Nov 2, 2011

### Rubiss

I calculated that Sy2=$\frac{\hbar^{2}}{4}$(1 0; 0 1). (That matrix is the identity matrix). I found that there is only one eigenvalue for this matrix: $\frac{\hbar^{2}}{4}$. Since this is the only eigenvalue of Sy2, this is the only value I could obtain if I were to measure Sy2. Is this correct? Have I done enough work, or do I need find the eigenstates of Sy2?

6. Nov 2, 2011

### vela

Staff Emeritus
Yup, that's right.